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Quy đồng mẫu thức các phân thức:
a. \({{25} \over {14{x^2}y}},{{14} \over {21x{y^5}}}\)
b. \({{11} \over {102{x^4}y}},{3 \over {34x{y^3}}}\)
c. \({{3x + 1} \over {12x{y^4}}},{{y – 2} \over {9{x^2}{y^3}}}\)
d. \({1 \over {6{x^3}{y^2}}},{{x + 1} \over {9{x^2}{y^4}}},{{x – 1} \over {4x{y^3}}}\)
e. \({{3 + 2x} \over {10{x^4}y}},{5 \over {8{x^2}{y^2}}},{2 \over {3x{y^5}}}\)
f. \({{4x – 4} \over {2x\left( {x + 3} \right)}},{{x – 3} \over {3x\left( {x + 1} \right)}}\)
g. \({{2x} \over {{{\left( {x + 2} \right)}^3}}},{{x – 2} \over {2x{{\left( {x + 2} \right)}^2}}}\)
h. \({5 \over {3{x^3} – 12x}},{3 \over {\left( {2x + 4} \right)\left( {x + 3} \right)}}\)
a. \({{14} \over {21x{y^5}}} = {2 \over {7x{y^5}}}\) MTC \( = 14{x^2}{y^5}\)
\({{14} \over {21x{y^5}}} = {2 \over {7x{y^5}}}\)\( = {{2.2x} \over {7x{y^5}.2x}} = {{4x} \over {14{x^2}{y^5}}}\); \({{25} \over {14{x^2}y}} = {{25.{y^4}} \over {14{x^2}y.{y^4}}} = {{25{y^4}} \over {14{x^2}{y^5}}}\)
b. MTC = \(102{x^4}{y^3}\)
\({{11} \over {102{x^4}y}} = {{11.{y^2}} \over {102{x^4}y.{y^2}}} = {{11{y^2}} \over {102{x^4}{y^3}}}\); \({3 \over {34x{y^3}}} = {{3.3{x^3}} \over {34x{y^3}.3{x^3}}} = {{9{x^3}} \over {102{x^4}{y^3}}}\)
c. MTC = \(36{x^3}{y^4}\)
\({{3x + 1} \over {12x{y^4}}} = {{\left( {3x + 1} \right).3x} \over {12x{y^4}.3x}} = {{9{x^2} + 3x} \over {36{x^2}{y^4}}}\); \({{y – 2} \over {9{x^2}{y^3}}} = {{\left( {y – 2} \right).4y} \over {9{x^2}{y^3}.4y}} = {{4{y^2} – 8y} \over {36{x^2}{y^4}}}\)
d. MTC = \(36{x^3}{y^4}\)
\({1 \over {6{x^3}{y^2}}} = {{1.6{y^2}} \over {6{x^3}{y^2}.6{y^2}}} = {{6{y^2}} \over {36{x^3}{y^4}}}\); \({{x + 1} \over {9{x^2}{y^4}}} = {{\left( {x + 1} \right).4x} \over {9{x^2}{y^4}.4x}} = {{4{x^2} + 4x} \over {36{x^3}{y^4}}}\)
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\({{x – 1} \over {4x{y^3}}} = {{\left( {x – 1} \right).9{x^2}y} \over {4x{y^3}.9{x^2}y}} = {{9{x^3}y – 9{x^2}y} \over {36{x^3}{y^4}}}\)
e. MTC = \(120{x^4}{y^5}\)
\({{3 + 2x} \over {10{x^4}y}} = {{\left( {3 + 2x} \right).12{y^4}} \over {10{x^4}y.12{y^4}}} = {{36{y^4} + 24x{y^4}} \over {120{x^4}{y^5}}}\)
\({5 \over {8{x^2}{y^2}}} = {{5.15{x^2}{y^3}} \over {8{x^2}{y^2}.15{x^2}{y^3}}} = {{75{x^2}{y^3}} \over {120{x^4}{y^5}}}\)
\({2 \over {3x{y^5}}} = {{2.40{x^3}} \over {3x{y^5}.40{x^3}}} = {{80{x^3}} \over {120{x^4}{y^5}}}\)
f. MTC = \(3x\left( {x + 3} \right)\left( {x + 1} \right)\) Vì \({{4x – 4} \over {2x\left( {x + 3} \right)}} = {{2\left( {x – 1} \right)} \over {x\left( {x + 3} \right)}}\)
\({{4x – 4} \over {2x\left( {x + 3} \right)}} = {{2\left( {x – 1} \right)} \over {x\left( {x + 3} \right)}} = {{2\left( {x – 1} \right).3\left( {x + 1} \right)} \over {x\left( {x + 3} \right).3\left( {x + 1} \right)}} = {{6\left( {{x^2} – 1} \right)} \over {3x\left( {x + 3} \right)\left( {x + 1} \right)}}\)
\({{x – 3} \over {3x\left( {x + 1} \right)}} = {{\left( {x – 3} \right)\left( {x + 3} \right)} \over {3x\left( {x + 1} \right)\left( {x + 3} \right)}} = {{{x^2} – 9} \over {3x\left( {x + 1} \right)\left( {x + 3} \right)}}\)
g. MTC = \(2x{\left( {x + 2} \right)^3}\)
\({{2x} \over {{{\left( {x + 2} \right)}^3}}} = {{2x.2x} \over {2x{{\left( {x + 2} \right)}^3}}} = {{4{x^2}} \over {2x{{\left( {x + 2} \right)}^3}}}\)
\({{x – 2} \over {2x{{\left( {x + 2} \right)}^2}}} = {{\left( {x – 2} \right)\left( {x + 2} \right)} \over {2x{{\left( {x + 2} \right)}^2}\left( {x + 2} \right)}} = {{{x^2} – 4} \over {2x{{\left( {x + 2} \right)}^3}}}\)
h. \(3{x^3} – 12x = 3x\left( {{x^2} – 4} \right) = 3x\left( {x – 2} \right)\left( {x + 2} \right)\)
\(\left( {2x + 4} \right)\left( {x + 3} \right) = 2\left( {x + 2} \right)\left( {x + 3} \right)\)
MTC = \(6x\left( {x – 2} \right)\left( {x + 2} \right)\left( {x + 3} \right)\)
\(\eqalign{ & {5 \over {3{x^3} – 12x}} = {5 \over {3x\left( {x – 2} \right)\left( {x + 2} \right)}} = {{5.2\left( {x + 3} \right)} \over {3x\left( {x – 2} \right)\left( {x + 2} \right).2\left( {x + 3} \right)}} \cr & = {{10\left( {x + 3} \right)} \over {6x\left( {x – 2} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} \cr & {3 \over {\left( {2x + 4} \right)\left( {x + 3} \right)}} = {3 \over {2\left( {x + 2} \right)\left( {x + 3} \right)}} = {{3.3x\left( {x – 2} \right)} \over {2\left( {x + 2} \right)\left( {x + 3} \right).3x\left( {x – 2} \right)}} \cr & = {{9x\left( {x – 2} \right)} \over {6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}} \cr} \)